The number of real values of `k` for which the lines `(x)/(1)=(y-1)/(k)=(z)/(-1) and (x-k)/(2k)=(y-k)/(3k-1)=(z-2)/(k)` are coplanar, is

Answer

To determine the number of real values of \( k \) for which the lines

\[ \frac{x}{1} = \frac{y-1}{k} = \frac{z}{-1} \]

and

\[ \frac{x-k}{2k} = \frac{y-k}{3k-1} = \frac{z-2}{k} \]

are coplanar, we can follow these steps:

### Step 1: Identify Points on the Lines
From the first line, we can identify a point on the line by setting the parameter to 0:
- For \( t = 0 \):
- \( x = 0 \)
- \( y = 1 \)
- \( z = 0 \)

Thus, the point \( A(0, 1, 0) \) lies on the first line.

From the second line, we can identify a point by setting the parameter to 0:
- For \( s = 0 \):
- \( x = k \)
- \( y = k \)
- \( z = 2 \)

Thus, the point \( B(k, k, 2) \) lies on the second line.

### Step 2: Find Direction Vectors
The direction vector of the first line can be derived from the coefficients:
- Direction vector \( \vec{d_1} = (1, k, -1) \)

The direction vector of the second line can be derived similarly:
- Direction vector \( \vec{d_2} = (2k, 3k-1, k) \)

### Step 3: Formulate the Condition for Coplanarity
The lines are coplanar if the scalar triple product of the vectors formed by points \( A \), \( B \), and the direction vectors \( \vec{d_1} \) and \( \vec{d_2} \) is zero.

The vector \( \vec{AB} \) from \( A \) to \( B \) is:
\[ \vec{AB} = (k - 0, k - 1, 2 - 0) = (k, k - 1, 2) \]

### Step 4: Set Up the Determinant
We can set up the determinant of the matrix formed by \( \vec{AB} \), \( \vec{d_1} \), and \( \vec{d_2} \):

\[ \begin{vmatrix} k & k - 1 & 2 \\ 1 & k & -1 \\ 2k & 3k - 1 & k \end{vmatrix} = 0 \]

### Step 5: Calculate the Determinant
Calculating the determinant, we get:

\[ = k \begin{vmatrix} k & -1 \\ 3k - 1 & k \end{vmatrix} - (k - 1) \begin{vmatrix} 1 & -1 \\ 2k & k \end{vmatrix} + 2 \begin{vmatrix} 1 & k \\ 2k & 3k - 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. \( \begin{vmatrix} k & -1 \\ 3k - 1 & k \end{vmatrix} = k^2 + 3k - 1 \)

2. \( \begin{vmatrix} 1 & -1 \\ 2k & k \end{vmatrix} = k + 2k = 3k \)

3. \( \begin{vmatrix} 1 & k \\ 2k & 3k - 1 \end{vmatrix} = 3k - 2k^2 - k = 2k - 2k^2 \)

Putting it all together:

\[ = k(k^2 + 3k - 1) - (k - 1)(3k) + 2(2k - 2k^2) = 0 \]

### Step 6: Simplify and Solve for \( k \)
Expanding and simplifying the equation leads to a cubic equation in \( k \):

\[ -k^3 + 4k^2 - 8k + 2 = 0 \]

### Step 7: Find the Number of Real Roots
To find the number of real values of \( k \), we can use the discriminant or numerical methods to find the roots of the cubic equation.

Using numerical methods or graphing, we can determine the number of real roots.

### Final Answer
The number of real values of \( k \) for which the lines are coplanar is **3**.